Question

$\underset{x\to 0}{lim}\frac{1-\cos x}{x(2^x-1)}$ is equal to

• A. $\frac{1}{2}{\log }_{2}e$
• B. $\frac{1}{2}{\log }_{e}2$
• C. $1$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}{\log }_{2}e$

$\underset{x\to 0}{lim}\frac{1-\cos x}{x(2^x-1)}$

Applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{\sin x}{-1+2^x+2^{x}x\log 2}$

Again applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{\cos x}{\left(\log 2\right)2^x(\left(\log 2\right)x+2)}$

$=\frac{1}{\log 4}=\frac{1}{2}{\log }_{2}e$